Simplify the following expression and state the condition under which the simplification is valid. You can assume that $x \neq 0$. $r = \dfrac{9}{6x^2 + 15x} \times \dfrac{2x + 5}{7} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ 9 \times (2x + 5) } { (6x^2 + 15x) \times 7 } $ $ r = \dfrac {9 (2x + 5)} {7 \times 3x(2x + 5)} $ $ r = \dfrac{9(2x + 5)}{21x(2x + 5)} $ We can cancel the $2x + 5$ so long as $2x + 5 \neq 0$ Therefore $x \neq -\dfrac{5}{2}$ $r = \dfrac{9 \cancel{(2x + 5})}{21x \cancel{(2x + 5)}} = \dfrac{9}{21x} = \dfrac{3}{7x} $